Integrand size = 23, antiderivative size = 72 \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {1}{2} a^2 (a+6 b) x+\frac {a^3 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 (3 a+b) \tanh (c+d x)}{d}-\frac {b^3 \tanh ^3(c+d x)}{3 d} \] Output:
1/2*a^2*(a+6*b)*x+1/2*a^3*cosh(d*x+c)*sinh(d*x+c)/d+b^2*(3*a+b)*tanh(d*x+c )/d-1/3*b^3*tanh(d*x+c)^3/d
Time = 1.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {6 a^2 (a+6 b) (c+d x)+3 a^3 \sinh (2 (c+d x))+4 b^2 \left (9 a+2 b+b \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{12 d} \] Input:
Integrate[Cosh[c + d*x]^2*(a + b*Sech[c + d*x]^2)^3,x]
Output:
(6*a^2*(a + 6*b)*(c + d*x) + 3*a^3*Sinh[2*(c + d*x)] + 4*b^2*(9*a + 2*b + b*Sech[c + d*x]^2)*Tanh[c + d*x])/(12*d)
Time = 0.32 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4634, 300, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (i c+i d x)^2\right )^3}{\sec (i c+i d x)^2}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {\left (-b \tanh ^2(c+d x)+a+b\right )^3}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (-\tanh ^2(c+d x) b^3+(3 a+b) b^2+\frac {a^2 (a+3 b)-3 a^2 b \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2}\right )d\tanh (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^3 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}+\frac {1}{2} a^2 (a+6 b) \text {arctanh}(\tanh (c+d x))+b^2 (3 a+b) \tanh (c+d x)-\frac {1}{3} b^3 \tanh ^3(c+d x)}{d}\) |
Input:
Int[Cosh[c + d*x]^2*(a + b*Sech[c + d*x]^2)^3,x]
Output:
((a^2*(a + 6*b)*ArcTanh[Tanh[c + d*x]])/2 + b^2*(3*a + b)*Tanh[c + d*x] - (b^3*Tanh[c + d*x]^3)/3 + (a^3*Tanh[c + d*x])/(2*(1 - Tanh[c + d*x]^2)))/d
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Time = 0.99 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {3 a^{3} \sinh \left (2 d x +2 c \right )+4 \left (b^{3} \operatorname {sech}\left (d x +c \right )^{2}+9 a \,b^{2}+2 b^{3}\right ) \tanh \left (d x +c \right )+6 a^{2} d x \left (a +6 b \right )}{12 d}\) | \(65\) |
derivativedivides | \(\frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \left (d x +c \right )+3 \tanh \left (d x +c \right ) a \,b^{2}+b^{3} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d}\) | \(77\) |
default | \(\frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \left (d x +c \right )+3 \tanh \left (d x +c \right ) a \,b^{2}+b^{3} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d}\) | \(77\) |
risch | \(\frac {a^{3} x}{2}+3 b \,a^{2} x +\frac {a^{3} {\mathrm e}^{2 d x +2 c}}{8 d}-\frac {a^{3} {\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {2 b^{2} \left (9 \,{\mathrm e}^{4 d x +4 c} a +18 a \,{\mathrm e}^{2 d x +2 c}+6 b \,{\mathrm e}^{2 d x +2 c}+9 a +2 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}\) | \(113\) |
Input:
int(cosh(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
Output:
1/12*(3*a^3*sinh(2*d*x+2*c)+4*(b^3*sech(d*x+c)^2+9*a*b^2+2*b^3)*tanh(d*x+c )+6*a^2*d*x*(a+6*b))/d
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (66) = 132\).
Time = 0.25 (sec) , antiderivative size = 270, normalized size of antiderivative = 3.75 \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {3 \, a^{3} \sinh \left (d x + c\right )^{5} - 4 \, {\left (18 \, a b^{2} + 4 \, b^{3} - 3 \, {\left (a^{3} + 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (18 \, a b^{2} + 4 \, b^{3} - 3 \, {\left (a^{3} + 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (30 \, a^{3} \cosh \left (d x + c\right )^{2} + 9 \, a^{3} + 72 \, a b^{2} + 16 \, b^{3}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (18 \, a b^{2} + 4 \, b^{3} - 3 \, {\left (a^{3} + 6 \, a^{2} b\right )} d x\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, a^{3} \cosh \left (d x + c\right )^{4} + 2 \, a^{3} + 24 \, a b^{2} + 16 \, b^{3} + {\left (9 \, a^{3} + 72 \, a b^{2} + 16 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \] Input:
integrate(cosh(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")
Output:
1/24*(3*a^3*sinh(d*x + c)^5 - 4*(18*a*b^2 + 4*b^3 - 3*(a^3 + 6*a^2*b)*d*x) *cosh(d*x + c)^3 - 12*(18*a*b^2 + 4*b^3 - 3*(a^3 + 6*a^2*b)*d*x)*cosh(d*x + c)*sinh(d*x + c)^2 + (30*a^3*cosh(d*x + c)^2 + 9*a^3 + 72*a*b^2 + 16*b^3 )*sinh(d*x + c)^3 - 12*(18*a*b^2 + 4*b^3 - 3*(a^3 + 6*a^2*b)*d*x)*cosh(d*x + c) + 3*(5*a^3*cosh(d*x + c)^4 + 2*a^3 + 24*a*b^2 + 16*b^3 + (9*a^3 + 72 *a*b^2 + 16*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d* cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(d*x + c))
Timed out. \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:
integrate(cosh(d*x+c)**2*(a+b*sech(d*x+c)**2)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (66) = 132\).
Time = 0.05 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.22 \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {1}{8} \, a^{3} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 3 \, a^{2} b x + \frac {4}{3} \, b^{3} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {6 \, a b^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \] Input:
integrate(cosh(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")
Output:
1/8*a^3*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) + 3*a^2*b*x + 4/3*b ^3*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6 *d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6* d*x - 6*c) + 1))) + 6*a*b^2/(d*(e^(-2*d*x - 2*c) + 1))
Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (66) = 132\).
Time = 0.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.11 \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {3 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, {\left (a^{3} + 6 \, a^{2} b\right )} {\left (d x + c\right )} - 3 \, {\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - \frac {16 \, {\left (9 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a b^{2} + 2 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \] Input:
integrate(cosh(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")
Output:
1/24*(3*a^3*e^(2*d*x + 2*c) + 12*(a^3 + 6*a^2*b)*(d*x + c) - 3*(2*a^3*e^(2 *d*x + 2*c) + 12*a^2*b*e^(2*d*x + 2*c) + a^3)*e^(-2*d*x - 2*c) - 16*(9*a*b ^2*e^(4*d*x + 4*c) + 18*a*b^2*e^(2*d*x + 2*c) + 6*b^3*e^(2*d*x + 2*c) + 9* a*b^2 + 2*b^3)/(e^(2*d*x + 2*c) + 1)^3)/d
Time = 0.14 (sec) , antiderivative size = 221, normalized size of antiderivative = 3.07 \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {a^2\,x\,\left (a+6\,b\right )}{2}-\frac {\frac {2\,a\,b^2}{d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,b^3+3\,a\,b^2\right )}{3\,d}+\frac {2\,a\,b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,\left (2\,b^3+3\,a\,b^2\right )}{3\,d}+\frac {2\,a\,b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {a^3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {2\,a\,b^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:
int(cosh(c + d*x)^2*(a + b/cosh(c + d*x)^2)^3,x)
Output:
(a^2*x*(a + 6*b))/2 - ((2*a*b^2)/d + (4*exp(2*c + 2*d*x)*(3*a*b^2 + 2*b^3) )/(3*d) + (2*a*b^2*exp(4*c + 4*d*x))/d)/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - ((2*(3*a*b^2 + 2*b^3))/(3*d) + (2*a*b^2*e xp(2*c + 2*d*x))/d)/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - (a^3*exp (- 2*c - 2*d*x))/(8*d) + (a^3*exp(2*c + 2*d*x))/(8*d) - (2*a*b^2)/(d*(exp( 2*c + 2*d*x) + 1))
Time = 0.22 (sec) , antiderivative size = 339, normalized size of antiderivative = 4.71 \[ \int \cosh ^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx=\frac {3 e^{10 d x +10 c} a^{3}+12 e^{8 d x +8 c} a^{3} d x +7 e^{8 d x +8 c} a^{3}+72 e^{8 d x +8 c} a^{2} b d x +48 e^{8 d x +8 c} a \,b^{2}+36 e^{6 d x +6 c} a^{3} d x +216 e^{6 d x +6 c} a^{2} b d x +36 e^{4 d x +4 c} a^{3} d x -12 e^{4 d x +4 c} a^{3}+216 e^{4 d x +4 c} a^{2} b d x -144 e^{4 d x +4 c} a \,b^{2}-96 e^{4 d x +4 c} b^{3}+12 e^{2 d x +2 c} a^{3} d x -11 e^{2 d x +2 c} a^{3}+72 e^{2 d x +2 c} a^{2} b d x -96 e^{2 d x +2 c} a \,b^{2}-32 e^{2 d x +2 c} b^{3}-3 a^{3}}{24 e^{2 d x +2 c} d \left (e^{6 d x +6 c}+3 e^{4 d x +4 c}+3 e^{2 d x +2 c}+1\right )} \] Input:
int(cosh(d*x+c)^2*(a+b*sech(d*x+c)^2)^3,x)
Output:
(3*e**(10*c + 10*d*x)*a**3 + 12*e**(8*c + 8*d*x)*a**3*d*x + 7*e**(8*c + 8* d*x)*a**3 + 72*e**(8*c + 8*d*x)*a**2*b*d*x + 48*e**(8*c + 8*d*x)*a*b**2 + 36*e**(6*c + 6*d*x)*a**3*d*x + 216*e**(6*c + 6*d*x)*a**2*b*d*x + 36*e**(4* c + 4*d*x)*a**3*d*x - 12*e**(4*c + 4*d*x)*a**3 + 216*e**(4*c + 4*d*x)*a**2 *b*d*x - 144*e**(4*c + 4*d*x)*a*b**2 - 96*e**(4*c + 4*d*x)*b**3 + 12*e**(2 *c + 2*d*x)*a**3*d*x - 11*e**(2*c + 2*d*x)*a**3 + 72*e**(2*c + 2*d*x)*a**2 *b*d*x - 96*e**(2*c + 2*d*x)*a*b**2 - 32*e**(2*c + 2*d*x)*b**3 - 3*a**3)/( 24*e**(2*c + 2*d*x)*d*(e**(6*c + 6*d*x) + 3*e**(4*c + 4*d*x) + 3*e**(2*c + 2*d*x) + 1))